If we denote the Cartesian coordinate of a $n$-dimensional Euclidean space by a $n$-vector $\mathbf{x}$, then a $n$-ball centered at the origin with radius $r$ is the set of points that satisfy

Taking this definition, a 1-ball is the line segment $[-r,r]$, which its “volume” is $2r$ and has no surface area. A 2-ball is a circle with radius $r$, which has the “surface” area of $2\pi r$ and “volume” of $\pi r^2$.

Let us denote the volume of an $n$-ball with radius $r$ by $V_n(r)$, and consider that $n$-ball is defined by

Obviously

Therefore, we can find $V_n(r)$ recursively:

## Proportionality of Volume

Firstly, we prove $V_n(r)$ is proportional to $r^n$ by induction:

Consider $n=1$, $V_1(r)=2r=rV_1(1)$. If we have $V_k(r)=r^kV_k(1)$, then

Thus

## Formula for Volume

If we denote for $n>1$,

then

and

valid for $n\ge 1$ if we specially define $I_0=1$, and which

where the Gamma function is defined as

with the property that $z\Gamma(z) = \Gamma(z+1)$, and $\Gamma(n)=(n-1)!$ for positive integers $n$. But we can also consider $I_n$ without resolving to the Gamma function. Consider that

It is easy to see that $I_1=\pi/2$, $I_2=4/3$, $I_3=3\pi/8$, and

thus

and

Therefore we have

Indeed, we can also prove by induction that

And the surface area of the $n$-ball is obtained by $\frac{d}{dr}V_n(r)$.

## Alternative Derivation

Below is an alternative derivation by Greg Huber (1982). “Gamma function derivation of n-sphere volumes”. Am. Math. Monthly 89(5):301–302 (PDF)

Consider

for the reason that $\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{x^2}{2\sigma^2}) du = 1$.

Denote the surface of $n$-ball by $\omega_{n-1}(r)$, we can write the volume as the integration of surface area as the radius expands from zero:

or the volume as the integration of all the points in the ball:

Consider the density integration on an infinitely large $n$-ball, where the density at point $\mathbf{x}$ is $e^{-r^2}$ with $r$ defined by

Using the first integral we have (assumed radius 1)

But using the second integral we have

Because they shall be equal, we have

We can also verify that $\frac{d}{dr}V_n(r) = \omega_{n-1}(r)$.

## Properties of n-ball

If we consider the $n$-ball with unit diameter instead of unit radius, its volume is $V_n(1/2)$. We can see that, due to the exponential nature in the denominator,

Therefore, the higher the dimension, the denser we can pack the balls. I first learned this result from Professor Raymond Yeung.